Question

Find the equation of the conic of which one focus lies at (2,1) one directrix is
x + y = 0 and it passes through (1,4) Also identify the conic and reduce the conic you obtained above to standard form.
Draw a rough sketch of the conic obtained above.

Answer 1

Answer on Question #78889 – Math – Analytic Geometry

Question

Find the equation of the conic of which one focus lies at (2,1) one directrix is $x + y = 0$ and it passes through (1,4). Also identify the conic and reduce the conic you obtained above to standard form. Draw a rough sketch of the conic obtained above.

Solution

Conic is defined as locus of a point moving in a plane such that the ratio of its distance from a fixed point $(F)$ to the fixed straight line is always a constant. This ratio is called as eccentricity.

The distance of point (1,4) from focus at (2,1) is

$$\sqrt{(1 - 2)^2 + (4 - 1)^2} = \sqrt{10}$$

The distance of point (1,4) from directrix $x + y = 0$ is

$$\frac{|1 + 4|}{\sqrt{(1)^2 + (1)^2}} = \frac{5\sqrt{2}}{2}$$

Find the eccentricity

$$e = \frac{\sqrt{10}}{\frac{5\sqrt{2}}{2}} = \frac{2\sqrt{5}}{5} < 1$$

Hence we have the ellipse.

The distance from an arbitrary point $(x,y)$ to the focus (2,1)

$$\sqrt{(x - 2)^2 + (y - 1)^2}$$

The distance of the point $(x,y)$ from directrix $x + y = 0$

$$\frac{|x + y|}{\sqrt{(1)^2 + (1)^2}} = \frac{|x + y|}{\sqrt{2}}$$

The eccentricity

$$e = \frac{\sqrt{(x - 2)^2 + (y - 1)^2}}{\frac{|x + y|}{\sqrt{2}}} = \frac{2\sqrt{5}}{5}$$ $$(x - 2)^2 + (y - 1)^2 = \frac{2}{5}(x + y)^2$$ $$x^2 - 4x + 4 + y^2 - 2y + 1 = \frac{2}{5}x^2 + \frac{4}{5}xy + \frac{2}{5}y^2$$ $$\frac{3}{5}x^2 - \frac{4}{5}xy + \frac{3}{5}y^2 - 4x - 2y + 5 = 0$$ $$3x^2 - 4xy + 3y^2 - 20x - 10y + 25 = 0$$$$\begin{array}{l} A = 3, B = -4, C = 3, D = -20, E = -2, F = 25 \\ \cot(2\theta) = \frac{A - C}{B} = \frac{3 - 3}{-4} = 0 \Rightarrow \theta = 45{}^{\circ} \\ \end{array}$$ $$\begin{array}{l} x = x' \cos \theta - y' \sin \theta \\ y = x' \sin \theta + y' \cos \theta \\ \end{array}$$ $$\begin{array}{l} x = x' \frac{\sqrt{2}}{2} - y' \frac{\sqrt{2}}{2} \\ y = x' \frac{\sqrt{2}}{2} + y' \frac{\sqrt{2}}{2} \\ 3 \left(x' \frac{\sqrt{2}}{2} - y' \frac{\sqrt{2}}{2}\right)^2 - 4 \left(x' \frac{\sqrt{2}}{2} - y' \frac{\sqrt{2}}{2}\right) \left(x' \frac{\sqrt{2}}{2} + y' \frac{\sqrt{2}}{2}\right) + \\ \end{array}$$ $$+ 3 \left(x ^ {\prime} \frac {\sqrt {2}}{2} + y ^ {\prime} \frac {\sqrt {2}}{2}\right) ^ {2} - 2 0 \left(x ^ {\prime} \frac {\sqrt {2}}{2} - y ^ {\prime} \frac {\sqrt {2}}{2}\right) - 1 0 \left(x ^ {\prime} \frac {\sqrt {2}}{2} + y ^ {\prime} \frac {\sqrt {2}}{2}\right) + 2 5 = 0$$ $$3 \left(\frac {(x ^ {\prime}) ^ {2}}{2} - x ^ {\prime} y ^ {\prime} + \frac {(y ^ {\prime}) ^ {2}}{2}\right) - 4 \left(\frac {(x ^ {\prime}) ^ {2}}{2} - \frac {(y ^ {\prime}) ^ {2}}{2}\right) + 3 \left(\frac {(x ^ {\prime}) ^ {2}}{2} + x ^ {\prime} y ^ {\prime} + \frac {(y ^ {\prime}) ^ {2}}{2}\right) -$$ $$- 1 0 \sqrt {2} x ^ {\prime} + 1 0 \sqrt {2} y ^ {\prime} - 5 \sqrt {2} x ^ {\prime} - 5 \sqrt {2} y ^ {\prime} + 2 5 = 0$$ $$(x ^ {\prime}) ^ {2} - 1 5 \sqrt {2} x ^ {\prime} + 5 (y ^ {\prime}) ^ {2} + 5 \sqrt {2} y ^ {\prime} + 2 5 = 0$$ $$(x ^ {\prime}) ^ {2} - 2 \left(\frac {1 5 \sqrt {2}}{2}\right) x ^ {\prime} + \frac {2 2 5}{2} - \frac {2 2 5}{2} + 5 \left((y ^ {\prime}) ^ {2} + 2 \left(\frac {5 \sqrt {2}}{2}\right) y ^ {\prime} + \frac {2 5}{2} - \frac {2 5}{2}\right) + 2 5 =$$ $$= 0$$ $$\left(x ^ {\prime} - \frac {1 5 \sqrt {2}}{2}\right) ^ {2} + 5 \left(y ^ {\prime} + \frac {5 \sqrt {2}}{2}\right) ^ {2} = 1 5 0$$ $$\frac {\left(x ^ {\prime} - \frac {1 5 \sqrt {2}}{2}\right) ^ {2}}{\left(5 \sqrt {6}\right) ^ {2}} + \frac {\left(y ^ {\prime} + \frac {5 \sqrt {2}}{2}\right) ^ {2}}{\left(\sqrt {3 0}\right) ^ {2}} = 1$$

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